"Reduce" function for Series

hlin117 picture hlin117 · Jan 26, 2016 · Viewed 17.4k times · Source

Is there an analog for reduce for a pandas Series?

For example, the analog for map is pd.Series.apply, but I can't find any analog for reduce.


My application is, I have a pandas Series of lists:

>>> business["categories"].head()

0                      ['Doctors', 'Health & Medical']
1                                        ['Nightlife']
2                 ['Active Life', 'Mini Golf', 'Golf']
3    ['Shopping', 'Home Services', 'Internet Servic...
4    ['Bars', 'American (New)', 'Nightlife', 'Loung...
Name: categories, dtype: object

I'd like to merge the Series of lists together using reduce, like so:

categories = reduce(lambda l1, l2: l1 + l2, categories)

but this takes a horrific time because merging two lists together is O(n) time in Python. I'm hoping that pd.Series has a vectorized way to perform this faster.

Answer

Mike Müller picture Mike Müller · Jan 26, 2016

With itertools.chain() on the values

This could be faster:

from itertools import chain
categories = list(chain.from_iterable(categories.values))

Performance

from functools import reduce
from itertools import chain

categories = pd.Series([['a', 'b'], ['c', 'd', 'e']] * 1000)

%timeit list(chain.from_iterable(categories.values))
1000 loops, best of 3: 231 µs per loop

%timeit list(chain(*categories.values.flat))
1000 loops, best of 3: 237 µs per loop

%timeit reduce(lambda l1, l2: l1 + l2, categories)
100 loops, best of 3: 15.8 ms per loop

For this data set the chaining is about 68x faster.

Vectorization?

Vectorization works when you have native NumPy data types (pandas uses NumPy for its data after all). Since we have lists in the Series already and want a list as result, it is rather unlikely that vectorization will speed things up. The conversion between standard Python objects and pandas/NumPy data types will likely eat up all the performance you might get from the vectorization. I made one attempt to vectorize the algorithm in another answer.